ISyE 3232C
Stochastic Manufacturing and Service Systems
Spring 2010
Practice Test 3 (Solution)
Closed book, closed notes, no electronics.
1. Let
X
p
be the service time for a perm, and
X
m
be the service time for a manicure. Then,
X
p
∼
exp(2)
and
X
m
∼
exp(3). Also, let
T
S
, T
L
, T
G
be the service time of Sue, Liz and Grace, respectively. Note
that
Y
L
=
X
p
, Y
G
=
X
m
and
Y
S
=
X
p
.
(a)
E
[min
{
Y
L
, Y
G
}
] +
E
[
Y
S
] =
1
2 + 3
+
1
2
=
7
10
(b)
P
(
Y
L
< Y
G
)
P
(
Y
S
> Y
G
) +
P
(
Y
L
> Y
G
)
P
(
Y
S
> Y
L
) =
2
2 + 3
3
2 + 3
+
3
2 + 3
2
2 + 2
=
27
50
(c)
P
(
Y
L
> Y
G
)
P
(
Y
L
> Y
S
) =
3
2 + 3
2
2 + 2
=
3
10
2. Let
X
A
be the remaining lifetime of person
A
, and
Y
be the interarrival time of kidney arrivals. Then,
X
A
∼
exp(
μ
A
) and
Y
∼
exp(
λ
). Then, the probability that A obtains a new kidney is same as
P
(
X
A
> Y
) =
λ
λ
+
μ
A
.
3.
{
N
(
t
) :
t
≥
0
}
is a P.P(
λ
= 30/hour).
(a)
E
[
S
4
] =
4
30
(b)
E
[
N
(7)

N
(5)

N
(4)

N
(3) = 3] =
E
[
N
(7)

N
(5)] =
E
[
N
(2)

N
(0)] =
E
[
N
(2)] = 2
×
30 = 60.
(c)
{
N
1
(
t
)
}
is a P.P(
λ
1
= 24/hour) of car arrivals, and
{
N
2
(
t
)
}
is a P.P(
λ
2
= 6/hour) of van arrivals.
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 Spring '07
 Billings
 Trigraph, xm, Imaginary number, service time

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